[数据结构与算法]树形数据结构返回前端示例代码

当前端需要返回如下结构的json时候：

{
    "success":true,
    "code":20000,
    "message":"成功",
    "data":{
        "list":[
            Object{...},
            Object{...},
            {
                "id":"1521703425235382274",
                "label":"数据库",
                "children":[
                    {
                        "id":"1521703425302491138",
                        "label":"mysql",
                        "children":[

                        ],
                        "pid":"1521703425235382274"
                    }
                ],
                "pid":"0"
            },
            Object{...}
        ]
    }
}

2种情况

1. 只有一颗树结构：
   例如一个系统有多个子系统

 @Test
    public void test03() {
        HashMap<Integer, TestDemo> map = new HashMap<>();
        List<TestDemo> list = new ArrayList<TestDemo>();
        list.add(new TestDemo(1, "1",0, new ArrayList<TestDemo>()));
        list.add(new TestDemo(2, "2", 1,new ArrayList<TestDemo>()));
        list.add(new TestDemo(3, "3", 2,new ArrayList<TestDemo>()));
        list.add(new TestDemo(4, "4", 1,new ArrayList<TestDemo>()));
        list.add(new TestDemo(5, "5", 4,new ArrayList<TestDemo>()));
        list.add(new TestDemo(6, "6", 4,new ArrayList<TestDemo>()));
        list.add(new TestDemo(9, "9", 2,new ArrayList<TestDemo>()));
        list.add(new TestDemo(7, "7", 3,new ArrayList<TestDemo>()));
        list.add(new TestDemo(8, "8", 3,new ArrayList<TestDemo>()));
        for (TestDemo testDemo : list) {
            map.put(testDemo.getKey(), testDemo);
        }
        TestDemo root = null;
        for (TestDemo testDemo : list) {
            if(testDemo.getPid() == 0){
            //寻找根节点
                root = testDemo;
                continue;
            }
            //父子关系构造，list之前new好了，所有这里直接取出来add就可以了。没有new的话，取值然后再去存值比较麻烦
            map.get(testDemo.getPid()).getChildren().add(testDemo);
        }

        System.out.println(root);

实体类

@Data
@AllArgsConstructor
@NoArgsConstructor
public class TestDemo {
    int key;
    String name;
    int pid;
    List<TestDemo>  children;
}

2.多颗树结构
例如多门学科为父根，每门学科有不同的章节

List<TestDemo> list = new ArrayList<TestDemo>();
        list.add(new TestDemo(2, "2", 1,new ArrayList<TestDemo>()));
        list.add(new TestDemo(3, "3", 2,new ArrayList<TestDemo>()));
        list.add(new TestDemo(4, "4", 1,new ArrayList<TestDemo>()));
        list.add(new TestDemo(5, "5", 4,new ArrayList<TestDemo>()));
        list.add(new TestDemo(6, "6", 4,new ArrayList<TestDemo>()));
        list.add(new TestDemo(1, "1",0, new ArrayList<TestDemo>()));
        list.add(new TestDemo(9, "9", 2,new ArrayList<TestDemo>()));
        list.add(new TestDemo(7, "7", 3,new ArrayList<TestDemo>()));
        list.add(new TestDemo(8, "8", 3,new ArrayList<TestDemo>()));
        //另一棵树，不同于上一棵
        list.add(new TestDemo(11, "11", 0,new ArrayList<TestDemo>()));
        list.add(new TestDemo(12, "12", 11,new ArrayList<TestDemo>()));
        list.add(new TestDemo(13, "13",11,new ArrayList<TestDemo>()));

以0有2棵树

增加 List roots = new ArrayList<>();
root = testDemo 改成 roots.add(testDemo);

其中的实体类中的list属性，个人建议new出来比较好。当后端查的数据和前端的数据不一致时，建议自己重新遍历生成一个vo对象包含初始化好list再去做父子关系构造

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